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Let $M$ be a closed manifold and $f:M\to M$ be a diffeomorphism. Suppose the homomorphism $f^*:H^k(M;\mathbb R)\to H^k(M;\mathbb R)$ has an eigenvalue $\lambda\in\mathbb{R}$. Note that $\lambda$ is algebraic but possibly not integer.

Does there always exist closed and not exact differential form $\alpha\in\Omega^k(M)$ such that $f_1^*(\alpha)=\lambda\alpha$ for some diffeomorphism $f_1$ homotopic to $f$?

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    $\begingroup$ I guess if you can find a Riemannian metric invariant under $f$ then you can use a harmonic form: en.wikipedia.org/wiki/De_Rham_cohomology#Harmonic_forms $\endgroup$
    – Qiaochu Yuan
    Commented Jul 2 at 22:26
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    $\begingroup$ Let's consider the case $M$ is a surface, $f$ is pseudo-Anosov, and $k=1$. A 1-form $\alpha \in \Omega^1(M)$ induces a singular foliation on $M$, and the statement that $f^*\alpha = \lambda \alpha$ means that that singular foliation is invariant under $f$. I somehow feel that the requirement that $\alpha$ be smooth will lead to a contradiction... $\endgroup$
    – André Henriques
    Commented Jul 2 at 22:50
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    $\begingroup$ @AndréHenriques the zeroes of $\alpha$ correspond to the singularities of this foliation. However, if foliation is given by a 1-form (not by quadratic differential), all its singularities have even number of prongs (not like in pseudo-Anosov theory). Finally, if the set of zeroes of $\alpha$ has nonempty interior, then $\alpha$ does not define a foliation on the whole of $M$ $\endgroup$
    – Andrey Ryabichev
    Commented Jul 2 at 23:32
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    $\begingroup$ @QiaochuYuan Unfortunately, a metric with nontrivial automorphism is an exceptional situation, so for arbitrary $M$ and $f$ such a metric does not exist $\endgroup$
    – Andrey Ryabichev
    Commented Jul 3 at 8:39
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    $\begingroup$ However, as you point out, $\alpha$ need not itself define a foliation. However, we can perturb $\alpha$ by a generic $1$-form to produce a foliation and thus a point in the Thurston boundary. If we send the perturbations to zero, do these points converge? A positive answer should give a negative answer to your question. But I'm not familiar enough with the Thurston boundary / simple closed curves to answer. $\endgroup$
    – Will Sawin
    Commented Jul 7 at 13:01

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